Ta có
\(\eqalign{
& a)\,\,{\kern 1pt} 0,6 + {2 \over { - 3}} = {6 \over {10}} + {{ - 2} \over 3} \cr
& = {3 \over 5} + {{ - 2} \over 3}{\kern 1pt} = {{3.3} \over {5.3}} + {{( - 2).5} \over {3.5}} \cr
& = {9 \over {15}} + {{ - 10} \over {15}} \cr
& = {{9 - 10} \over {15}} = {{ - 1} \over {15}} \cr} \)
\(\eqalign{
& b)\,\,{1 \over 3} - \left( { - 0,4} \right) = {1 \over 3} + 0,4 \cr
& = {1 \over 3} + {4 \over {10}} = {1 \over 3} + {2 \over 5} \cr
& = {{1.5} \over {3.5}} + {{2.3} \over {5.3}} = {5 \over {15}} + {6 \over {15}} \cr
& = {{5 + 6} \over {15}} = {{11} \over {15}} \cr} \)
